Question: How many pairs of positive integers $(x,y)$ satisfy $x^2-y^2=51$?
Solution: Factoring, we have $(x+y)(x-y)=51$.  Since $x,y$ are positive, we have $x+y>x-y>0$.  Note that $51=51*1=17*3$.  Thus either $x+y=51$, $x-y=1$ or $x+y=17$, $x-y=3$.  Solving in the first case gives $x=26,y=25$, and the second case gives $x=10,y=7$.  Thus there are $\boxed{2}$ pairs $(x,y)$ that solve the equation.